I've attached the question as an image here:

For a) I get 9 square units

and b) I get (243pi)/5 cube units

Help would be much appreciated

a) is correct

I'd be interested in seeing your work for (b). I don't know how you arrived at that answer.

Consider a infinitesimal cross section of the solid between $(x,x+dx)$

It's a circular disk of radius $\sqrt{x}$ and thickness $dx$

Thus it's differential volume is

$dV = \pi (\sqrt{x})^2~dx = \pi x~dx$

$V = \displaystyle \int_0^9~dV = \displaystyle \int_0^9~\pi x = \left . \pi \dfrac{x^2}{2} \right|_0^9 = \dfrac{81\pi}{2}$

Hi Romsek,

Thanks for the reply! Unfortunately I'm confused by the method you've shown as my level of maths is not very advanced as of yet. I'll upload an image of my working using the method I've been shown. Perhaps you could point out what I've done incorrectly?

Originally Posted by Quoctopus
Hi Romsek,

Thanks for the reply! Unfortunately I'm confused by the method you've shown as my level of maths is not very advanced as of yet. I'll upload an image of my working using the method I've been shown. Perhaps you could point out what I've done incorrectly?

Bah, my answer was when the initial curve is rotated about the x-axis.

Let me redo this and get back to you.

Ok.

Consider a cross section of the volume at height $z$ with thickness $dz$

This circular disk has radius $r=z^2$ and $z \in [0,3]$

$dV = \pi r^2~dz = \pi z^4~dz$

$V = \displaystyle \int_0^3 dV = \int_0^3 \pi z^4 ~dz = \dfrac{243\pi}{5}$

So you were right all along. My apologies.